, 2023

a) Denote $x_i, y_i$ as the roots of the $i$-th polynomial, then by Vieta's theorem, $a_i = -(x_i + y_i)$ and $b_i = x_i y_i$. Thus $a_i b_i = -x_i y_i (x_i + y_i)$ which is always even, implying that at most 1 number among $a_i, b_i$ is odd. Hence, there are at most 10 odd values among their coefficients.

The equality case occurs when $(1, 2), (3, 4), \dots, (9, 10), (-1, -2), \dots, (-9, -10)$ are roots of the given polynomials.

b) By Vieta's theorem, we need to find the minimum and maximum value of
$$T = x_1 y_1 + x_2 y_2 + \dots + x_{10} y_{10}.$$
Note that for all $x, y \in \mathbb{R}$, $xy \ge -\frac{x^2 + y^2}{2}$, thus
$$T \ge -\frac{1}{2}(x_1^2 + y_1^2 + x_2^2 + y_2^2 + \dots + x_{10}^2 + y_{10}^2) = -(1^2 + 2^2 + \dots + 10^2) = -385.$$
On the other hand, for $x, y \in \mathbb{Z}$ and $x \ne y$ then $(x-y)^2 \ge 1$ so $xy \le \frac{x^2 + y^2 - 1}{2}$, thus
$$T \le \frac{1}{2}(x_1^2 + y_1^2 + x_2^2 + y_2^2 + \dots + x_{10}^2 + y_{10}^2) - 5 = 380.$$
Hence, we can conclude that:

* $\max T = 380$, attained when $(1, 2), (3, 4), \dots, (9, 10), (-1, -2), \dots, (-9, -10)$ are roots of 10 polynomials.
* $\min T = -385$, attained when $(1, -1), (2, -2), \dots, (10, -10)$ are roots of 10 polynomials.